$L=\underset{x\to 0}{L\mathrm{im}} \frac{\sqrt{{\displaystyle \frac{\cos 2x+{\left(1+3x\right)}^{1/3}}{2}}}-\sqrt[3]{{\displaystyle \frac{4 {\cos }^{3}x-\ln {\left(1+x\right)}^4}{4}}}}{x}$If $L=a/b$, where '$a$‘ and ’$b$' are relative primes, find $\frac{(a+b)}{19}$

Denote $L_1=\underset{x\to 0}{L\mathrm{im}} \frac{\sqrt{{\displaystyle \frac{\cos 2x+{\left(1+3x\right)}^{1/3}}{2}}}}{x}$ and $L_2=\underset{x\to 0}{L\mathrm{im}} \frac{\sqrt[3]{{\displaystyle \frac{4 {\cos }^{3}x-\ln {\left(1+x\right)}^4}{4}}}-1}{x}$

We have $L=L_1-L_2$

Now for $L_1$ on rationalization

$$\begin{gathered} L_1=\underset{x\to 0}{L\mathrm{im}} \frac{{\displaystyle \frac{\cos 2x+{\left(1+3x\right)}^{1/3}}{2}}}{x·\left(\sqrt{\frac{\cos 2x+{\left(1+3x\right)}^{1/3}}{2}}+1\right)} \\ {L}_1=\frac{1}{2}·\underset{x\to 0}{L\mathrm{im}} \frac{\left[{\left(1+3x\right)}^{1/3}-1\right]-\left[1-\cos 2x\right]}{2x}=\frac{1}{4}·\left[\underset{x\to 0}{L\mathrm{im}} \frac{{\left(1+3x\right)}^{1/3}-1}{x}-\underset{x\to 0}{L\mathrm{im}} \underset{zero}{\underbrace{\frac{1-\cos 2x}{x}}}\right] \\ =\frac{1}{4}·\left[\underset{x\to 0}{L\mathrm{im}} \frac{(1+x+.....-1)}{x}-0\right]=\frac{1}{4} \end{gathered}$$

For $L_2=\underset{x\to 0}{L\mathrm{im}} \frac{{\left[{\displaystyle \frac{4 {\cos }^{3}x-\ln {\left(1+x\right)}^4}{4}}\right]}^{1/3}-1}{x}=\underset{x\to 0}{L\mathrm{im}} \frac{{\left[{\cos }^{3}x-\ln (1+x)\right]}^{1/3}-1}{x}$

$=\frac{1}{3}· \underset{x\to 0}{L\mathrm{im}} \frac{{\cos }^{3}x-\ln (1+x)-1}{x}$             (Using $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$)

Now $L_2=\frac{1}{3}·\left[\underset{x\to 0}{L\mathrm{im}} \frac{\left(\cos x-1\right)}{x}·\underset{x\to 0}{L\mathrm{im}}\left({\cos }^{2}x+1+\cos x\right)-\underset{x\to 0}{L\mathrm{im}}\frac{\ln (1+x)}{x}\right]$

$$\begin{gathered} =\frac{1}{3}\left[0-1\right]=-\frac{1}{3} \\ L=\frac{1}{4}+\frac{1}{3}=\frac{7}{12} \Rightarrow a+b=19 \end{gathered}$$